Determine an equation of the line that is perpendicular to the line 3x + 12y = 1 and contains the point (-2, 6)

Step-by-step explanation:

find 3x + 12y = 1 in terms of y

so y = -3/12 z + 1/3

y = -1/4 x + 1/3

gradient of the line(m) = -1/4

gradient of the line perpendicular to the line 3x + 12y = 1 is -1/m = -1 ÷ -1/4 = -1 x -4 = 4

(OR change the sign to its opposite(- to + and + to -) then put the denominator in place of the nominator and the nominator in place of the denominator)

for this line x1=-2 and y1=6 and m=4

so using the equation

(y-y1) = m(x-x1)

y-6 = 4(x-(-2)

y-6 = 4(x+2)

y-6 = 4x + 8

y = 4x + 14

i dont know

step-by-step explanation: